The Stacks project

Lemma 38.30.2. Let $R$ be a ring and let $f \in R$. Let $r\geq 0$ be an integer. Let $R \to S$ be a ring map and let $M$ be an $S$-module. Assume

  1. $R \to S$ is of finite presentation and flat,

  2. every fibre ring $S \otimes _ R \kappa (\mathfrak p)$ is geometrically integral over $R$,

  3. $M$ is a finite $S$-module,

  4. $M_ f$ is a finitely presented $S_ f$-module,

  5. for all $\mathfrak p \in R$, $f \not\in \mathfrak p$ with $\mathfrak q = \mathfrak pS$ the module $M_{\mathfrak q}$ is free of rank $r$ over $S_\mathfrak q$.

Then there exists a finitely generated ideal $I \subset R$ with $V(f) = V(I)$ such that for all $a \in I$ with $R' = R[\frac{I}{a}]$ the quotient

\[ M' = (M \otimes _ R R')/a\text{-power torsion} \]

over $S' = S \otimes _ R R'$ satisfies the following: for every prime $\mathfrak p' \subset R'$ there exists a $g \in S'$, $g \not\in \mathfrak p'S'$ such that $M'_ g$ is a free $S'_ g$-module of rank $r$.

Proof. This lemma is a generalization of More on Algebra, Lemma 15.26.5; we urge the reader to read that proof first. Choose a surjection $S^{\oplus n} \to M$, which is possible by (1). Choose a finite submodule $K \subset \mathop{\mathrm{Ker}}(S^{\oplus n} \to M)$ such that $S^{\oplus n}/K \to M$ becomes an isomorphism after inverting $f$. This is possible by (4). Set $M_1 = S^{\oplus n}/K$ and suppose we can prove the lemma for $M_1$. Say $I \subset R$ is the corresponding ideal. Then for $a \in I$ the map

\[ M_1' = (M_1 \otimes _ R R')/a\text{-power torsion} \longrightarrow M' = (M \otimes _ R R')/a\text{-power torsion} \]

is surjective. It is also an isomorphism after inverting $a$ in $R'$ as $R'_ a = R_ f$, see Algebra, Lemma 10.70.7. But $a$ is a nonzerodivisor on $M'_1$, whence the displayed map is an isomorphism. Thus it suffices to prove the lemma in case $M$ is a finitely presented $S$-module.

Assume $M$ is a finitely presented $S$-module satisfying (3). Then $J = \text{Fit}_ r(M) \subset S$ is a finitely generated ideal. By Lemma 38.9.3 we can write $S$ as a direct summand of a free $R$-module: $\bigoplus _{\alpha \in A} R = S \oplus C$. For any element $h \in S$ writing $h = \sum a_\alpha $ in the decomposition above, we say that the $a_\alpha $ are the coefficients of $h$. Let $I' \subset R$ be the ideal of coefficients of elements of $J$. Multiplication by an element of $S$ defines an $R$-linear map $S \to S$, hence $I'$ is generated by the coefficients of the generators of $J$, i.e., $I'$ is a finitely generated ideal. We claim that $I = fI'$ works.

We first check that $V(f) = V(I)$. The inclusion $V(f) \subset V(I)$ is clear. Conversely, if $f \not\in \mathfrak p$, then $\mathfrak q = \mathfrak p S$ is not an element of $V(J)$ by property (5) and More on Algebra, Lemma 15.8.7. Hence there is an element of $J$ which does not map to zero in $S \otimes _ R \kappa (\mathfrak p)$. Thus there exists an element of $I'$ which is not contained in $\mathfrak p$, so $\mathfrak p \not\in V(fI') = V(I)$.

Let $a \in I$ and set $R' = R[\frac{I}{a}]$. We may write $a = fa'$ for some $a' \in I'$. By Algebra, Lemmas 10.70.2 and 10.70.8 we see that $I' R' = a'R'$ and $a'$ is a nonzerodivisor in $R'$. Set $S' = S \otimes _ S R'$. Every element $g$ of $JS' = \text{Fit}_ r(M \otimes _ S S')$ can be written as $g = \sum _\alpha c_\alpha $ for some $c_\alpha \in I'R'$. Since $I'R' = a'R'$ we can write $c_\alpha = a'c'_\alpha $ for some $c'_\alpha \in R'$ and $g = (\sum c'_\alpha )a' = g' a'$ in $S'$. Moreover, there is an $g_0 \in J$ such that $a' = c_\alpha $ for some $\alpha $. For this element we have $g_0 = g'_0 a'$ in $S'$ where $g'_0$ is a unit in $S'$. Let $\mathfrak p' \subset R'$ be a prime ideal and $\mathfrak q' = \mathfrak p'S'$. By the above we see that $JS'_{\mathfrak q'}$ is the principal ideal generated by the nonzerodivisor $a'$. It follows from More on Algebra, Lemma 15.8.9 that $M'_{\mathfrak q'}$ can be generated by $r$ elements. Since $M'$ is finite, there exist $m_1, \ldots , m_ r \in M'$ and $g \in S'$, $g \not\in \mathfrak q'$ such that the corresponding map $(S')^{\oplus r} \to M'$ becomes surjective after inverting $g$.

Finally, consider the ideal $J' = \text{Fit}_{r - 1}(M')$. Note that $J'S'_ g$ is generated by the coefficients of relations between $m_1, \ldots , m_ r$ (compatibility of Fitting ideal with base change). Thus it suffices to show that $J' = 0$, see More on Algebra, Lemma 15.8.8. Since $R'_ a = R_ f$ (Algebra, Lemma 10.70.7) and $M'_ a = M_ f$ we see from (5) that $J'_ a$ maps to zero in $S_{\mathfrak q''}$ for any prime $\mathfrak q'' \subset S'$ of the form $\mathfrak q'' = \mathfrak p''S'$ where $\mathfrak p'' \subset R'_ a$. Since $S'_ a \subset \prod _{\mathfrak q''\text{ as above}} S'_{\mathfrak q''}$ (as $(S'_ a)_{\mathfrak p''} \subset S'_{\mathfrak q''}$ by Lemma 38.7.4) we see that $J'R'_ a = 0$. Since $a$ is a nonzerodivisor in $R'$ we conclude that $J' = 0$ and we win. $\square$

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